Description : The area of a trapezium is 475 cm2 and the height is 19 cm. -Maths 9th
Last Answer : Let the length of one parallel side of trapezium is a. So length of other parallel side = a + 4 Given the area of trapezium = 475 & the height = 19 Now area of a trapezium = 21 height ... =23∴Length of smalle rparallel side of trapezium =23 & Length of other parallel side of trapezium = 23+4=27
Last Answer : According to question find the lengths of its two parallel sides
Description : In trapezium ABCD, AB|| DC and diagonals AC and BD intersect at O. If area of triangle AOD is 30cm square , find the area of triangle BOC -Maths 9th
Last Answer : In the given figure: Area of triangle ADC = Area of triangle BCD (Triangles on the same and parallel) Now subtract the area of triangle DOC from both of them so... (Area of triangle ADC - Area of ... => Area of triangle AOD = Area of triangle BOC Hence the area of triangle BOC is 30 cm square.
Description : Find the area of a trapezium whose parallel sides are 25 cm and 13 cm long and the distance between them is 8 cm. -Maths 9th
Last Answer : Area of a trapezium = 1/2(sum of parallel sides) x (perpendicular diatance between them) = 1/2(25 + 13) x 8 = 152cm2
Description : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that AB||HC || GD || FE. Also BC=CD=DE, and GF=6 cm... -Maths 9th
Last Answer : Given : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in the fig. These pieces are arranged ... length of coloured tape required = 30 cm (b) The values are : Happiness, beauty, Knowledge.
Description : ABCD is a trapezium with parallel sides AB = a cm and DC = b cm . E and F are the mid - points of the non - parallel sides . -Maths 9th
Last Answer : Clearly, EF = AB + DC / 2 = a + b / 2 Let h be the height , then ar(Trap. ABFE) : ar(Trap. EFCD) ⇒ 1/2 [a + (a+b / 2)] × h : 1/2 [b + (a+b / 2)] × h ⇒ 2a + a + b / 2 : 2b + a + b / 2 ⇒ 3a + b : 3b + a
Description : ABCD is a trapezium with parallel sides AB = a cm and DC = b cm. -Maths 9th
Last Answer : Solution of this question
Description : ABCD is trapezium in which AB || DC, DC = 30 cm and AB = 50 cm. -Maths 9th
Last Answer : According to question prove that ar (DCYX) = 7/9 ar (XYBA).
Description : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that ... the card by putting up a colored tape on the non-parallel sides of the trapezium.. -Maths 9th
Last Answer : Let us consider the following lay out of the greeting card. Trapeziums are arranged in such a way that AB || HC || GD || FE. Also BC=CD=DE and GF=6 cm and DE = 4cm. If three parallel lines make equal ... HG+GF+BC+CD+DE = 6+6+6+4+4+4=30 cm. (b) The values are: Happiness, beauty, Knowledge.
Description : The base of a right prism is a trapezium. The lengths of the parallel sides are 8 cm and 14 cm and the distance between the parallel -Maths 9th
Last Answer : Area of trapezium =12×h(AB+CD) =12×8×(8+14)=12×8×(8+14) =4×22=88cm2=4×22=88cm2 = Volume of prism = Height of prism ×× area of base ⇒height×88=1056 (given)⇒height×88=1056 (given) ⇒height×88=105688⇒height×88=105688 ⇒12cm =12×h(AB+CD)
Description : In a trapezium ABCD, AB is parallel to CD, BD is perpendicular to AD. AC is perpendicular to BC. If AD = BC = 15 cm and AB = 25 cm, -Maths 9th
Last Answer : answer:
Description : A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field. -Maths 9th
Last Answer : Let the given field is in the form of a trapezium ABCD such that parallel sides are AB = 10 m and DC = 25 m Non-parallel sides are AD = 13 m and BC = 14 m. We draw BE || AD, such that BE = 13 m. ... = 112 m2 So, area of the field = area of ∆BCE + area of parallelogram ABED = 84 m2 + 112 m2 = 196 m2
Description : Find the area of the trapezium PQRS with height PQ given in the figure given below. -Maths 9th
Last Answer : The area of trapezium =
Description : How to make Working model on the area of trapezium -Maths 9th
Last Answer : NEED ANSWER
Description : working model on area of trapezium -Maths 9th
Description : a working model on area of trapezium -Maths 9th
Last Answer : This answer was deleted by our moderators...
Description : What is the area of trapezium ? -Maths 9th
Last Answer : Area of trapezium= 1/2(sum of parallel sides)×height
Description : The area of a trapezium is 39cm2 . The distance between its parallel sides is 6cm. If one of the parallel sides is 5cm, then find the other parallel side. -Maths 9th
Last Answer : area of trapezium = 39 cm sq height of trpezium = 6 cm one || side = 5 cm let other be x area of trapezium = 1/2 × ( sum of || sides) × h 39 = 1/2 × ( 5+x) × 6 39×2 = (5+x) × 6 78/6 = 5+x 13-5 = x 8 = x x = 8cm
Description : Area of trapezium -Maths 9th
Last Answer : 1/2*(sum of parallel sides)*ht
Description : ABCD is a trapezium with AB and CD as parallel sides. The diagonals intersect at O. The area of the triangle ABO is p and that of triangle CDO is q. -Maths 9th
Description : Find the radius of a sphere whose surface area is 154 cm square. -Maths 9th
Last Answer : Let 'r' be the radius of sphere Surface area of sphere = 4 πr2 ⇒ 154 = 4 πr2 ⇒ 154 = 4 x 22/7 x r2 ⇒ r 2 = 154 x 7/4 x 22 = 49/4 ⇒ r = 7/2 cm = 3.5 cm
Description : A solid cylinder has total surface area of 462 cm square. -Maths 9th
Last Answer : Let r cm be the radius of the base and h cm be the height of the cylinder, Then, total surface area of cylinder = 2 πr (r + h) Curved surface area of cylinder = 2 πrh We have, Curved surface area = 1/3(Total surface ... x 22 = 7/2 cm Volume of the cylinder = πr2h = 22/7 x 7 x 7 x 7/2 = 539 cm3
Description : The area of the curved surface and the area of the base of a right circular cylinder are a square cm and b square cm respectively -Maths 9th
Description : The curved surface of a cylinder is developed into a square whose diagonal is 2√2 cm. The area of the base of the cylinder (in cm^2) is -Maths 9th
Description : 4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC. -Maths 9th
Last Answer : . Solution: Given that, ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. To prove, F is the mid-point of BC. Proof, BD intersected EF at G. In ΔBAD, E is the ... point of BD and also GF || AB || DC. Thus, F is the mid point of BC (Converse of mid point theorem)
Description : ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ΔABC ≅ ΔBAD (iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.] -Maths 9th
Last Answer : ] Solution: To Construct: Draw a line through C parallel to DA intersecting AB produced at E. (i) CE = AD (Opposite sides of a parallelogram) AD = BC (Given) , BC = CE ⇒∠CBE = ∠CEB also, ∠A+∠CBE = ... BC (Given) , ΔABC ≅ ΔBAD [SAS congruency] (iv) Diagonal AC = diagonal BD by CPCT as ΔABC ≅ ΔBA.
Description : ABCD is a trapezium in which AB II CD and AD = BC (see flg). Show that: -Maths 9th
Last Answer : Given: ABCD is a trapezium, in which AB || DC and AD = BC. To Prove: (i) ∠A = ∠B (ii) ∠C = ∠D (iii) △ABC ≅ △BAD (iv) Diagonal AC = diagonal BD. Const.: Produce AB to E, such that a line through ... △ABC ≅ △BAD [by SAS congruence axiom] (iv) ⇒ AC = BD [c.p.c.t.] Thus, diagonal AC = diagonal BD.
Description : E is the mid-point of the side AD of the trapezium ABCD with AB || DC. -Maths 9th
Last Answer : Given ABCD is a trapezium in which AB || DC and EF||AB|| CD. Construction Join, the diagonal AC which intersects EF at O. To show F is the mid-point of BC. Proof Now, in ΔADC, E is the mid-point of AD ... 0 is the mid-point of AC and OF || AB. So, by mid-point theorem, F is the mid-point of BC.
Description : E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. -Maths 9th
Last Answer : According to question the mid-points of the non-parallel sides AD and BC of a trapezium ABCD.
Description : Prove that the line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides of the trapezium. -Maths 9th
Last Answer : Given Let ABCD be a trapezium in which AB|| DC and let M and N be the mid-points of the diagonals AC and BD, respectively.
Description : In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th
Last Answer : According to question prove that ar (ABCD) = ar (APQD).
Description : If non-parallel sides of a trapezium are equal, prove that it is cyclic. -Maths 9th
Last Answer : Given ABCD is a trapezium whose non-parallel sides AD and BC are equal.
Description : A field is in the shape of a trapezium having parallel sides 90 m and 30 m. -Maths 9th
Last Answer : In trapezium ABCD, we draw a perpendicular line CE to the line AB.
Description : Prove that the line segment joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides and equal to half of their difference. -Maths 9th
Last Answer : In a parallelogram ABCD, the bisector of ∠ A also bisects BC at X.Prove that AD = 2AB.
Description : E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that EF||AB and EF = 1/2 (AB +CD). -Maths 9th
Last Answer : Solution :-